On the gap between prime ideals

1.   Introduction

In the rational number field $\mathbb{Q}$, we say two prime numbers $p, q$ have gap $k$ if $p-q = \pm k$ for a positive integer $k$. Two prime numbers are called twin primes if and only if their gap is 2. Studying the pair of prime numbers of a given gap is a main subject in analytic number theory. In this paper, we generalize the definition of gap to a number field.

First, we define the gap function $G_{K}$ in a number field $K$.

Definition 1.1. Let $K$ be a number field, $\mathcal{O}_{K}$ the ring of integers in $K$, and Spec$(\mathcal{O}_{K})$ the set of prime ideals of $\mathcal{O}_{K}$. The gap function $G_{K}$ for $K$ is the map:

for $\mathfrak{p}, \mathfrak{q}\in{\rm Spec}(\mathcal{O}_{K})$, $\mathbb{N}\mathfrak{p} = (\mathcal{O}_{K}:\mathfrak{p})$ is the norm of $\mathfrak{p}$.

Then we consider the following questions:

Question 1: Given a number field $K$ and a positive number $N$, do there exist infinitely many pairs of prime ideals such that their gap $G_{K} = N$?

Question 1 can be difficult even though $K = \mathbb{Q}$. For example, Question 1 becomes the twin prime conjecture when $K = \mathbb{Q}$ and $N = 2$. To study the above question, we first need to study the following question:

Question 2: Given a number field $K$ and a positive number $N$, do there exist two distinct prime ideals such that their gap $G_{K} = N$?

Note that if $K = \mathbb{Q}, N = 1$, the only pair of prime numbers satisfying $G_{\mathbb{Q}} = 1$ is $(2, 3)$. However, when $K$ is a number field and $K\neq\mathbb{Q}$, we will show that there may not exist two prime ideals whose gap is $1$.

In this paper, we study a very special case of Question 2: $K$ is a quadratic number field or cyclotomic field and $N = 1$. Moreover, we show that if $K$ is a quadratic number field or cyclotomic field that has two distinct prime ideals of gap 1, then the number of such pairs of prime ideals is finite. Thus we give an answer to the Question 1 for a special case.

In particular, we prove the following two results:

Theorem 1.2. Let $d\neq1$ be a square-free integer, and $K$ the quadratic field $\mathbb{Q}[\sqrt{d}]$. Then the pairs of prime ideals $\mathfrak{p}, \mathfrak{q}$ with gap 1 in $K$ are as follows: let $\mathfrak{p}\cap\mathbb{Z} = (p), \mathfrak{q}\cap\mathbb{Z} = (q)$,

1. $p = 3$ and $q = 2$, where $d\equiv1, 3, 6, 7, 9, 10 \;{\rm mod}\; 12$;

2. $p = 5$ and $q = 2$, where $d\equiv5, 21, 29 \;{\rm mod}\; 40$.

Theorem 1.3. Let $N$ be a positive integer $(N\neq1, 2, 3, 4, 6)$, $K$ the cyclotomic field $\mathbb{Q}(\zeta_{N})$, and $\mathcal{O}_{K}$ the ring of integers of $K$. Then there exist two distinct prime ideals $\mathfrak{p}, \mathfrak{q}$ of $\mathcal{O}_{K}$ such that $G_{K}(\mathfrak{p}, \mathfrak{q}) = 1$ if and only if $N = q, 2q$, where $q$ is a Mersenne prime number and $q\neq3$.

The proofs of above results are given in section 3. The principal method here is to calculate the decomposition of the prime number in the quadratic field and the cyclotomic field. In section 4 we give several examples. Section 2 is a summary of some results from algebraic number theory which are used throughout the paper.

2.   Preliminaries

In this section, we review all the facts we need in the rest of the paper. To study the gap between prime ideals for quadratic fields and cyclotomic fields, we need to know how the prime number $p$ decomposes in their ring of integers.

Lemma 2.1. Let $d\neq1$ be a square-free integer, $K$ the quadratic field $\mathbb{Q}[\sqrt{d}]$, $disc(\mathcal{O}_{K}/\mathbb{Z})$ be the discriminant of $K$, and p an odd prime number. Then we have the following results.

1. If $p$ divides $disc(\mathcal{O}_{K}/\mathbb{Z})$, then (p) ramifies in $\mathcal{O}_{K}$.

2. For p not dividing the d, we have

(p) is the product the two distinct ideals if and only if d is a square $\rm mod$ p.

(p) is a prime ideals in $\mathbb{Q}[\sqrt{d}]$ if and only if d is not a square $\rm mod$ p.

3. For the prime number 2 when $d\equiv 1 \;{\rm mod}\; 4$, we have

(2) is the product the two distinct ideals in $\mathbb{Q}[\sqrt{d}]$ if and only if $d\equiv 1 \;{\rm mod}\; 8$.

(2) is a prime ideals in $\mathbb{Q}[\sqrt{d}]$ if and only if $d\equiv 5 \;{\rm mod}\; 8$.

The proof of Lemma 2.1 can be found in ^[1].

Lemma 2.2. Let $N = \Pi_{p}p^{v_{p}}$ be the prime factorization of the positive integer $N$, and let $f_{p}$ be the smallest positive integer such that

Then one has in the cyclotomic field $\mathbb{Q}(\zeta_{N})$ the factorization

where $\mathfrak{p}_{1}, \mathfrak{p}_{2}, ...\mathfrak{p}_{r}$ are distinct prime ideals, all of degree $f_{p}$.

The proof of Lemma 2.2 can be found in ^[2]. The following result helps us treat the case when $K$ is a cyclotomic field. And the proof can be found in ^[3] and ^[4].

Theorem 2.3 (Mihǎilescu). The only solutions of the equation

in integers $a, b\geq2$ and non-zero integers $x, y$ are given by $(\pm3)^{2}-2^{3} = 1$.

3.   Proof of the main results

3.1.   Proof of Theorem 1.2

Proof. Let $\mathfrak{p}, \mathfrak{q}$ be two distinct prime ideals in $K = \mathbb{Q}[\sqrt{d}]$. Then $\mathfrak{p}\cap\mathbb{Z} = p\mathbb{Z}, \mathfrak{q}\cap\mathbb{Z} = q\mathbb{Z}$, where $p, q$ are two prime numbers in $\mathbb{Z}$. Now we have

where $f_{p} = (\mathcal{O}_{K}:\mathfrak{p}), f_{q} = (\mathcal{O}_{K}:\mathfrak{q})$ are the degree of residue fields. Note that the degree of field extension $[\mathbb{Q}[\sqrt{d}]:\mathbb{Q}] = 2$ is divided by $f_{p}$ and $f_{q}$, then $f_{p}, f_{q}$ can only be $1$ or $2$. We suppose that $G_{K}(\mathfrak{p}, \mathfrak{q}) = p^{f_{p}}-q^{f_{q}} = 1$. Then

Let $\cdot\overwithdelims()p$ denotes the Legendre symbol. We list all possible cases:

1. $p = 3, q = 2;f_{p} = f_{q} = 1$

    (a) Both $2$ and $3$ split in $\mathbb{Q}[\sqrt{d}]$. By Lemma 2.1, we have

Therefore $d \equiv 1 \;{\rm mod}\; 24$.

    (b) 2 splits in $\mathbb{Q}[\sqrt{d}]$, 3 ramifies in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 9 \;{\rm mod}\; 24$.

    (c) Both 2 and 3 ramify in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 3, 6 \;{\rm mod}\; 12$.

    (d) 2 ramifies in $\mathbb{Q}[\sqrt{d}]$, 3 splits in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 7, 10 \;{\rm mod}\; 12$.

2. $p = 2, q = 3;f_{p} = 2, f_{q} = 1$

    (a) 2 is a prime ideal in $\mathbb{Q}[\sqrt{d}]$, 3 ramifies in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 21 \;{\rm mod}\; 24$.

    (b) 2 is a prime ideal in $\mathbb{Q}[\sqrt{d}]$, 3 splits in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 13 \;{\rm mod}\; 24$.

3. $p = 5, q = 2;f_{p} = 1, f_{q} = 2$

    (a) 2 is a prime ideal in $\mathbb{Q}[\sqrt{d}]$, 5 ramifies in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 5 \;{\rm mod}\; 40$.

    (b) 2 is a prime ideal in $\mathbb{Q}[\sqrt{d}]$, 5 splits in $\mathbb{Q}[\sqrt{d}]$. Then

Therefore, $d \equiv 21, 29 \;{\rm mod}\; 40$.

In summary, $d\equiv1, 3, 6, 7, 9, 10\; {\rm mod}\; 12, or d\equiv5, 21, 29 \; {\rm mod}\; 40$.

3.2.   Proof of Theorem 1.3

Proof. Let $N = \Pi_{p}p^{v_{p}}$ be the prime factorization of the positive integer $N$, $N\neq1, 2, 3, 4, 6$. And let $\mathfrak{p}, \mathfrak{q}$ be two distinct prime ideals in $K = \mathbb{Q}(\zeta_{N})$. Then $\mathfrak{p}\cap\mathbb{Z} = p\mathbb{Z}, \mathfrak{q}\cap\mathbb{Z} = q\mathbb{Z}$, where $p, q$ are two prime numbers in $\mathbb{Z}$. We suppose that:

where $f_{p}, f_{q}$ are the degree of residue fields.

1. $f_{p}, f_{q} > 1$: By Theorem 2.3, $p = 3, f_{3} = 2; q = 2, f_{2} = 3$. Lemma 2.2 implies that

In other words, $\frac{N}{3^{v_{3}}}|8$, $\frac{N}{2^{v_{2}}}|7$. The second equation implies that $N = 2^{v_{2}}\cdot7$ or $2^{v_{2}}$.

    (a) If $N = 2^{v_{2}}$: then $\frac{N}{3^{v_{3}}}|8$ implies that $v_{3} = 0$ and $N = 2^{v_{2}}, v_{2}\leq3$. Then we have $v_{2} = 1$, $N = 2$, $K = \mathbb{Q}$; $v_{2} = 2$, $N = 4$, $K = \mathbb{Q}[\sqrt{-1}]$; $v_{2} = 3$, $N = 8$, $K = \mathbb{Q}(\zeta_{8})$. However, when $K = \mathbb{Q}(\zeta_{8})$, $2$ totally ramifies in $K$, $f_{2} = 1\neq 3$. So $K = \mathbb{Q}(\zeta_{8})$ is impossible.

    (b) If $N = 7\cdot2^{v_{2}}$: then $\frac{N}{3^{v_{3}}} = 7\cdot2^{v_{2}}$ doesn't divide $8$. So $N = 7\cdot2^{v_{2}}$ is impossible. Therefore, $f_{p}, f_{q} > 1$ is impossible.

2. $f_{p} = 1$: (3.1) becomes $p-q^{f_{q}} = 1$. Now $q^{f_{q}}+1 = p$ is an odd prime number. Then $q = 2$, $f_{2} = 2^{n}$ for some nonnegative integer $n$ and $p$ is a Fermat prime number. Lemma 2.2 implies that

In other words, $\frac{N}{p^{v_{p}}}|(p-1)$ and $\frac{N}2^{v_{2}}|(p-2)$. Therefore, $N = 2^{v_{2}}$, 2 totally ramifies in $K$ and $f_{2} = 2^{2^{n}} = 1$. From $n = 0$, we obtain $p = 2^{2^{n}}+1 = 2+1 = 3$, and $2^{v_{2}}|2$. So $N = 1, 2$.

Therefore, $f_{p} = 1$ is impossible.

3. $f_{q} = 1$: (3.1) becomes $2^{f_{2}}-q = 1$. Now $q = 2^{f_{2}}-1$ is a Mersenne prime number. Again, Lemma 2.2 implies that

From the first equation, we obtain $N = 2^{v_{2}}q^{v_{q}}$. The second equation shows that $2^{v_{2}}|(2^{f_{2}}-2)$. Note that $f_{2}\neq1$, otherwise $v_{q} = 0, v_{2} = 0, 1$ and $N = 1, 2$. And $2^{f_{2}}-2 = 2(2^{f_{2}-1}-1)$, $2^{f_{2}-1}-1$ is an odd number. Then

    (a) If $v_{2} = 0$: $N|(2^{f_{2}}-1)$ implies that $N = q$. Now we show that $f_{2}$ is indeed the smallest positive integer such that

If there is a positive integer $d$, satisfying $d|f_{2}$ and $2^{d} \equiv 1 \;{\rm mod}\; q$, then $2^{d} < 2^{f_{2}} = q+1$, which leads to a contradiction. Therefore, $N = q$.

    (b) If $v_{2} = 1$: $\frac{N}{2}|(2^{f_{2}}-1)$ implies that $N = 2q$. Similarly, $f_{2}$ is indeed the smallest positive integer such that

Therefore, $N = 2q$.

In summary, there exist two distinct prime ideals $\mathfrak{p}, \mathfrak{q}$ of $\mathcal{O}_{K}$ such that $G_{K}(\mathfrak{p}, \mathfrak{q}) = 1$ if and only if $N = q, 2q$, where $q$ is a Mersenne prime number and $q\neq3$.

Remark 3.1. Let $K = \mathbb{Q}(\zeta_{q})$ or $\mathbb{Q}(\zeta_{2q})$, where $q$ is a Mersenne prime not equal to $3$. According to the proof of Theorem 1.3, if $\mathfrak{p}$ and $\mathfrak{q}$ satisfy $G_{K}(\mathfrak{p}, \mathfrak{q}) = 1$, then $\mathfrak{p}\cap\mathbb{Z} = 2\mathbb{Z}, \mathfrak{q}\cap\mathbb{Z} = q\mathbb{Z}$, and $q = 2^{f_{2}}-1$ is a Mersenne prime. And let $\varphi$ be the Euler's totient function. By Lemma 2.2, we have the following decomposition in $\mathcal{O}_{K}$

where $g$ is determined by

Let $A_{\mathfrak{p}, \mathfrak{q}}^{K}$ be the set of pairs of prime ideals of gap 1, which is

Then the above discussion shows that $\#A_{\mathfrak{p}, \mathfrak{q}}^{K} = g = \frac{q-1}{f_{2}}$.

4.   Examples

In this section, we give some examples of cyclotomic fields satisfying the Theorem 1.3. Recall that a Mersenne prime is of the form $M_{n} = 2^{n}-1$ for some positive integer $n$. For $n > 2$, the smallest Mersenne prime is $2^{3}-1 = 7$. Then we obtain the following example:

Example 4.1. Let $K = \mathbb{Q}[\zeta_{7}]$. According to Theorem 1.3, there exists two distinct prime ideals $\mathfrak{p}, \mathfrak{q}$ such that $G_{K}(\mathfrak{p}, \mathfrak{q}) = 1$, where $\mathfrak{p}\cap\mathbb{Z} = 2\mathbb{Z}, \mathfrak{q}\cap\mathbb{Z} = 7\mathbb{Z}$. The degree of residue field $f_{2}$ is the order of $2$ in $(\mathbb{Z}/7\mathbb{Z})^{\times}$, that is $f_{2} = 3$. Therefore, $2$ decomposes into $\frac{\varphi(7)}{f_{2}} = 2$ distinct prime ideals $\mathfrak{p}_{1}, \mathfrak{p}_{2}$ in $\mathcal{O}_{K}$. Moreover, $7$ ramifies in $\mathcal{O}_{K}$ with ramification index $e_{7} = \varphi(7) = 6$. Explicitly,

Example 4.2. $M_{7} = 2^{13}-1 = 8191$ is a Mersenne prime. According to Theorem 1.3, $K = \mathbb{Q}[\zeta_{2\cdot8191}]$ has two distinct prime ideals $\mathfrak{p}, \mathfrak{q}$ such that $G_{K}(\mathfrak{p}, \mathfrak{q}) = 1$, where $\mathfrak{p}\cap\mathbb{Z} = 2\mathbb{Z}, \mathfrak{q}\cap\mathbb{Z} = 8191\mathbb{Z}$. 2 decomposes into $\frac{\varphi(2\cdot8191)}{f_{2}} = \frac{8190}{13} = 630$ distinct prime ideals $\mathfrak{p}_{1}, \mathfrak{p}_{2}, ..., \mathfrak{p}_{630}$ in $\mathcal{O}_{K}$. $8191$ ramifies in $\mathcal{O}_{K}$ with ramification index $e_{8191} = \varphi(8191) = 8190$. Explicitly,

Acknowledgments

The author would like to thank referees for their careful corrections to and valuable comments on the original version of this paper.

Conflict of interest

The authors declare no conflict of interest.